Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1;
const int A = 10;
const int L = 18;
int n, m, q;
struct Graph {
  int par[N];
  int hgt[N];
  int anc[N][L];
  vector<int> ten[N][L];
  vector<int> con[N];
  vector<int> adj[N];
} graph;
vector<int> operator+(vector<int> &a, vector<int> &b) {
  vector<int> res(A);
  int k = 0;
  for (int i = 0, j = 0; k < A && (i < a.size() || j < b.size()); ++k) {
    if (i < a.size() && j < b.size() && a[i] == b[j]) {
      i++;
    }
    if (j == b.size() || i < a.size() && a[i] < b[j]) {
      res[k] = a[i++];
    } else {
      res[k] = b[j++];
    }
  }
  res.resize(k);
  return res;
}
void operator+=(vector<int> &a, vector<int> &b) { a = a + b; }
void dfs_lca(int u, int h) {
  sort(graph.con[u].begin(), graph.con[u].end());
  if (graph.con[u].size() > A) {
    graph.con[u].resize(A);
  }
  graph.hgt[u] = h;
  graph.anc[u][0] = graph.par[u];
  graph.ten[u][0] = graph.con[graph.par[u]];
  for (int i = 1; i < L; ++i) {
    graph.anc[u][i] = graph.anc[graph.anc[u][i - 1]][i - 1];
    graph.ten[u][i] =
        graph.ten[u][i - 1] + graph.ten[graph.anc[u][i - 1]][i - 1];
  }
  for (int v : graph.adj[u]) {
    if (v != graph.par[u]) {
      graph.par[v] = u;
      dfs_lca(v, h + 1);
    }
  }
}
vector<int> lca(int u, int v) {
  vector<int> a = graph.con[u];
  vector<int> b = graph.con[v];
  if (graph.hgt[u] < graph.hgt[v]) {
    swap(u, v);
  }
  int dif = graph.hgt[u] - graph.hgt[v];
  for (int i = L - 1; i >= 0; --i) {
    if ((dif >> i) & 1) {
      a += graph.ten[u][i];
      u = graph.anc[u][i];
    }
  }
  if (u == v) {
    return a + b;
  }
  for (int i = L - 1; i >= 0; --i) {
    if (graph.anc[u][i] != graph.anc[v][i]) {
      a += graph.ten[u][i];
      b += graph.ten[v][i];
      u = graph.anc[u][i];
      v = graph.anc[v][i];
    }
  }
  a += graph.con[graph.par[u]];
  return a + b;
}
void input() {
  cin >> n >> m >> q;
  for (int i = 1; i < n; ++i) {
    int u;
    cin >> u;
    int v;
    cin >> v;
    graph.adj[u].push_back(v);
    graph.adj[v].push_back(u);
  }
  for (int i = 1; i <= m; ++i) {
    int u;
    cin >> u;
    graph.con[u].push_back(i);
  }
}
void solve() {
  dfs_lca(1, 1);
  for (int i = 0; i < q; ++i) {
    int u;
    cin >> u;
    int v;
    cin >> v;
    int a;
    cin >> a;
    vector<int> res = lca(u, v);
    cout << min((int)res.size(), a) << " ";
    for (int j = 0; j < min((int)res.size(), a); ++j) {
      cout << res[j] << " ";
    }
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  input();
  solve();
}